Question

the height of the Empire State Building is 318.00 meters. if a stone is dropped from the top of the building, what is the stone's velocity just before is strikes the ground?
a. -1.75 meters/sec
b. -18 meters/sec
c. -79 meters/sec
d. -636 meters/sec

Answer 1

\[\Delta y=v_{0}t+\frac{1}{2}at^2\]\[v^2=v_{0}^2+2a \Delta y\]\[v=v_{0}+at\]\[\Delta y=\frac{v_{0}+v}{2}t\]now we have these kinematic equations that will give us what we need. now look at what you have there in the question. you have the building's height and we know that it's the Empire State Building that we're dropping things off of so we know that we're on Earth. That means the stone's acceleration is going to be the the gravity constant of -9.8m/s^2. There's one more thing... the stone is dropped. so that means it started at rest which also means that it's initial velocity is going to be zero. now what equation has these three variables were where can solve for the final velocity?